Let g be a generator of the multiplicative group of the field; assuming x and y are nonzero, we can write x=g^{a} and y=g^{b} with 0 <= a,b < p^{2}-1, and then x^{p+1}=y^{4} becomes g^{a(p+1)}=g^{4b}, or equivalently a(p+1) = 4b (mod p^{2}-1).

From this we see that p+1 | 4b is necessary, and if 4b=k(p+1) then (a,b) gives a solution iff a=k (mod p-1). Since a can range from 0 to p^{2}-2, then, there are either 0 solutions or p+1 solutions for any fixed b. The total number of nonzero solutions is therefore (p+1)* #{b | p+1 divides 4b}, and then (x,y)=(0,0) is the remaining solution.

Now if p is 1 (mod 4) we have p+1 | 4b iff b is a multiple of (p+1)/2, and there are 2(p-1) such b up to p^{2}-1, so there are 2(p-1)(p+1)+1 = 2p^{2}-1 solutions.

On the other hand, if p is 3 (mod 4) then p+1 | 4b iff b is a multiple of (p+1)/4, so we have 4(p-1) such b and there are 4p^{2}-3 solutions.